Q11RP
Question
The initial value problem \[\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = 3}}{{\bf{x}}^{\frac{{\bf{2}}}{{\bf{3}}}}}{\bf{,}}\;{\bf{x(0) = 1}}\] has a unique solution in some open interval around t = 0.
Step-by-Step Solution
Verified Answer
The given statement is true.
1Step 1: Finding partial derivatives
Since \[\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = 3}}{{\bf{x}}^{\frac{{\bf{2}}}{{\bf{3}}}}}\]
Then\[\frac{{\partial {\bf{f}}}}{{\partial {\bf{x}}}}{\bf{ = }}\frac{{\bf{2}}}{{\sqrt[{\bf{3}}]{{\bf{x}}}}}\]
2Step 2: Checking the final result
Apply the initial conditions\[x\left( 0 \right) = 1\]
\[\frac{{\partial f}}{{\partial x}}{\bf{ = }}\frac{{\bf{2}}}{{\bf{0}}}\]
The result is infinite.
The given function is discontinuous a x = 0. So the function is not continuous in a rectangle containing the point (0,1).
Therefore, the statement is True.
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